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$A$ is a $m \times m$ diagonal-constant matrix with $1$ on the main diagonal and $2$ on the first superdiagonal. all other elements are equal to $0$.

$$A= \begin{bmatrix} 1& 2& 0& 0& 0&0& ...&0 \\ 0& 1& 2& 0& 0& 0& ...& 0\\0&0&1&2&...\\...&&&&&&&2\\0&0&0&...&&&&1 \end{bmatrix}$$

And we know:

$$A^{-1} = \begin{bmatrix} 1& -2& 4& &&&&-2^{m-1} \\0& 1& -2& 4& &&&-2^{m-2}\\ &&\\&&&\\\\0&0&0&&&1&-2&4\\0&0&0&&&&1&-2\\0&0&0&...&&&&1 \end{bmatrix}$$

I understand the eigenvalues for matrix $A$ is $1$. It seems we can give a nontrivial upper bound on $\sigma_{m}$, $m^{th}$ singular value, knowing $A^{-1}$. Does any one know how?

Thanks.

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    $\begingroup$ None that I know of uses the ideas you're trying to apply here. Of course, it's possible to get an upper bound, but I'm not sure how strong you need the upper bound to be. An easy way to get an upper bound is to compute the Frobenius norm, but that uses nothing about the inverse. $\endgroup$ – Omnomnomnom Aug 9 '16 at 2:37
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    $\begingroup$ Well, now this makes much more sense. In fact, we have $\sigma_m(A) = 1/\sigma_1(A^{-1})$. Perhaps that is of use to you. $\endgroup$ – Omnomnomnom Aug 9 '16 at 3:07
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    $\begingroup$ One way is to use the definitions $$ \sigma_1(A) = \max \frac{\|Ax\|}{\|x\|}, \qquad \sigma_m(A) = \min \frac{\|Ax\|}{\|x\|} $$ and make the substitution $x = A^{-1}y$. $\endgroup$ – Omnomnomnom Aug 9 '16 at 3:15
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    $\begingroup$ Although the mystery of $m$ is solved, still one prevails: what is $\sigma_m$? $\endgroup$ – Algebraic Pavel Aug 14 '16 at 0:41
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    $\begingroup$ @Math4Life If it helps I think I had a typo. $\sigma_m(A) = \frac{1}{\sigma_1(A^{-1})}$ $\endgroup$ – Crimson Sep 21 '17 at 3:19

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